Determine the molar solubility of pbso4 in pure water. ksp (pbso4 = 1.82 × 10-8.

PbSO4 -> Pb2+ + SO4 2-

So there are 2 different particles. 1 mol of PbSO4 produces 1 mol of Pb2+, and 1 mol of SO4 2- (one of each).

Kps = [Pb2+] [SO4 2-]

We will call “s” the molarity of PbSO4

So the molarity of Pb2+ is also “s” (same number)

And the molarity of SO4 2- is also “s” (same number)

Kps = s*s = s2

1.82 × 10-8 = s2

So, in order to find s, we have to make the square root of 1.82 × 10-8, which is: 0,00013491 M

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